Question 12.6: (a) Prove properties 12.31, 12.32, 12.33, and 12.34. (b) Sho...

(a) Equation 12.31: As in Problem 12.4(a), $\hat{\rho}^{\dagger}=\left(\sum_{k} p_{k}\left|\Psi_{k}\right\rangle\left\langle\Psi_{k}\right|\right)^{\dagger}=\sum_{k}\left|\Psi_{k}\right\rangle\left\langle\Psi_{k}\right| p_{k}^{*}$, Since $p_k$ is real,

$\rho^{\dagger}=\rho$           (12.31).

$\hat{\rho}^{\dagger}=\hat{\rho}$.

Equation 12.32:

$\operatorname{Tr}(\rho)=1$           (12.32).

$\operatorname{Tr}[\rho]=\sum_{i} \rho_{i i}=\sum_{i} \sum_{k} p_{k}\left\langle e_{i} \mid \Psi_{k}\right\rangle\left\langle\Psi_{k} \mid e_{i}\right\rangle=\sum_{k} p_{k}\left\langle\Psi_{k}\left|\sum_{i}\right| e_{i}\right\rangle\left\langle e_{i}|| \Psi_{k}\right\rangle$.

$=\sum_{k} p_{k}\left\langle\Psi_{k} \mid \Psi_{k}\right\rangle=\sum_{k} p_{k}=1$.

In the last two steps I used the fact that each wave function is normalized, and Equation (12.30).

$0 \leq p_{k} \leq 1 \text { and } \sum_{k} p_{k}=1$                   (12.30).

Equation 12.33:

$\langle A\rangle=\operatorname{Tr}(\rho A )$           (12.33).

$\operatorname{Tr}(\rho A )=\sum_{i}(\rho A )_{i i}=\sum_{i} \sum_{j} \rho_{i j} A_{j i}=\sum_{i} \sum_{j} \sum_{k} p_{k}\left\langle e_{i} \mid \Psi_{k}\right\rangle\left\langle\Psi_{k} \mid e_{j}\right\rangle\left\langle e_{j}|\hat{A}| e_{i}\right\rangle$.

$=\sum_{k} p_{k}\left\langle\Psi_{k}\left|\left(\sum_{j}\left|e_{j}\right\rangle\left\langle e_{j}\right|\right) \hat{A}\left(\sum_{i}\left|e_{i}\right\rangle\left\langle e_{i}\right|\right)\right| \Psi_{k}\right\rangle=\sum_{k} p_{k}\left\langle\Psi_{k}|\hat{A}| \Psi_{k}\right\rangle=\langle A\rangle$.

Equation 12.34: As in Problem 12.4(b):

$i \hbar \frac{d \hat{\rho}}{d t}=[\hat{H}, \hat{\rho}] , \text { (if } \frac{d p_{k}}{d t}=0 \text { for all } k \text { ) }$           (12.34).

$i \hbar \frac{d}{d t} \hat{\rho}=i \hbar \sum_{k} p_{k}\left(\left|\dot{\Psi}_{k}\right\rangle\left\langle\Psi_{k}|+| \Psi_{k}\right\rangle\left\langle\dot{\Psi}_{k}\right|\right)=\sum_{k} p_{k}\left(\hat{H}\left|\Psi_{k}\right\rangle\left\langle\Psi_{k}|-| \Psi_{k}\right\rangle\left\langle\Psi_{k}\right| \hat{H}\right)$.

$=\hat{H} \hat{\rho}-\hat{\rho} \hat{H}=[\hat{H}, \hat{\rho}]$.

(b)

$\operatorname{Tr}\left(\rho^{2}\right)=\sum_{i}\left(\rho^{2}\right)_{i i}=\sum_{i} \sum_{k} \sum_{j} p_{k} p_{j}\left\langle e_{i} \mid \Psi_{k}\right\rangle\left\langle\Psi_{k} \mid \Psi_{j}\right\rangle\left\langle\Psi_{j} \mid e_{i}\right\rangle$.

$=\sum_{k} \sum_{j} p_{k} p_{j}\left\langle\Psi_{j}\left|\left(\sum_{i}\left|e_{i}\right\rangle\left\langle e_{i}\right|\right)\right| \Psi_{k}\right\rangle\left\langle\Psi_{k} \mid \Psi_{j}\right\rangle=\sum_{k} \sum_{j} p_{k} p_{j}\left|\left\langle\Psi_{k} \mid \Psi_{j}\right\rangle\right|^{2}$.

The wave functions are normalized, so $\left|\left\langle\Psi_{k} \mid \Psi_{j}\right\rangle\right| \leq 1$, with equality if and only if k = j. Therefore, unless this is a pure state

$\operatorname{Tr}\left(\rho^{2}\right)<\sum_{k} \sum_{j} p_{k} p_{j}=\sum_{k} p_{k} \sum_{j} p_{j}=1$ . QED

(c) We already know that $\rho^{2}=\rho$ for a pure state. In part (b) we proved that $\operatorname{Tr}\left(\rho^{2}\right)<1$ for a non-pure state, whereas from (a) Tr(ρ) = 1 for any density matrix. Therefore, for a non-pure state $\rho^{2} \neq \rho$ (they have different traces). QED